3.2.0 ∫ (a+b arctan(cx))2 _________________________

Integrand size = 25, antiderivative size = 273 \[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx=-\frac {b c (a+b \arctan (c x))}{d x}-\frac {3 c^2 (a+b \arctan (c x))^2}{2 d}-\frac {(a+b \arctan (c x))^2}{2 d x^2}+\frac {i c (a+b \arctan (c x))^2}{d x}+\frac {b^2 c^2 \log (x)}{d}-\frac {b^2 c^2 \log \left (1+c^2 x^2\right )}{2 d}-\frac {2 i b c^2 (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )}{d}-\frac {c^2 (a+b \arctan (c x))^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {b^2 c^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )}{d}-\frac {i b c^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{d}-\frac {b^2 c^2 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d} \]

-b*c*(a+b*arctan(c*x))/d/x-3/2*c^2*(a+b*arctan(c*x))^2/d-1/2*(a+b*arctan(c*x))^2/d/x^2+I*c*(a+b*arctan(c*x))^2

/d/x+b^2*c^2*ln(x)/d-1/2*b^2*c^2*ln(c^2*x^2+1)/d-2*I*b*c^2*(a+b*arctan(c*x))*ln(2-2/(1-I*c*x))/d-c^2*(a+b*arct

an(c*x))^2*ln(2-2/(1+I*c*x))/d-b^2*c^2*polylog(2,-1+2/(1-I*c*x))/d-I*b*c^2*(a+b*arctan(c*x))*polylog(2,-1+2/(1

+I*c*x))/d-1/2*b^2*c^2*polylog(3,-1+2/(1+I*c*x))/d

Rubi [A] (veriﬁed)

Time = 0.46 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {4990, 4946, 5038, 272, 36, 29, 31, 5004, 5044, 4988, 2497, 5114, 6745} \[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx=-\frac {i b c^2 \operatorname {PolyLog}\left (2,\frac {2}{i c x+1}-1\right ) (a+b \arctan (c x))}{d}-\frac {3 c^2 (a+b \arctan (c x))^2}{2 d}-\frac {2 i b c^2 \log \left (2-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))}{d}-\frac {c^2 \log \left (2-\frac {2}{1+i c x}\right ) (a+b \arctan (c x))^2}{d}-\frac {(a+b \arctan (c x))^2}{2 d x^2}+\frac {i c (a+b \arctan (c x))^2}{d x}-\frac {b c (a+b \arctan (c x))}{d x}-\frac {b^2 c^2 \operatorname {PolyLog}\left (2,\frac {2}{1-i c x}-1\right )}{d}-\frac {b^2 c^2 \operatorname {PolyLog}\left (3,\frac {2}{i c x+1}-1\right )}{2 d}-\frac {b^2 c^2 \log \left (c^2 x^2+1\right )}{2 d}+\frac {b^2 c^2 \log (x)}{d} \]

Int[(a + b*ArcTan[c*x])^2/(x^3*(d + I*c*d*x)),x]

-((b*c*(a + b*ArcTan[c*x]))/(d*x)) - (3*c^2*(a + b*ArcTan[c*x])^2)/(2*d) - (a + b*ArcTan[c*x])^2/(2*d*x^2) + (

I*c*(a + b*ArcTan[c*x])^2)/(d*x) + (b^2*c^2*Log[x])/d - (b^2*c^2*Log[1 + c^2*x^2])/(2*d) - ((2*I)*b*c^2*(a + b

*ArcTan[c*x])*Log[2 - 2/(1 - I*c*x)])/d - (c^2*(a + b*ArcTan[c*x])^2*Log[2 - 2/(1 + I*c*x)])/d - (b^2*c^2*Poly

Log[2, -1 + 2/(1 - I*c*x)])/d - (I*b*c^2*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)])/d - (b^2*c^2*Poly

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -

Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a

+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u

], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^

n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),

x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[(a + b*ArcTan[c*x])

^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Dist[b*c*(p/d), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2 - 2/(1 + e*(x/d))

]/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[1/d, I

nt[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f), Int[(f*x)^(m + 1)*((a + b*ArcTan[c*x])^p/(d + e*x)),

x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0] && LtQ[m, -1]

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p +

1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,

Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), Int[(f*x)^(m + 2)*((a + b*ArcTan[c*x])^p/(d + e*x

^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(-I)*((a + b*ArcT

an[c*x])^(p + 1)/(b*d*(p + 1))), x] + Dist[I/d, Int[(a + b*ArcTan[c*x])^p/(x*(I + c*x)), x], x] /; FreeQ[{a, b

, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p, 0]

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*(a + b*Ar

cTan[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)), x] + Dist[b*p*(I/2), Int[(a + b*ArcTan[c*x])^(p - 1)*(PolyLog[2, 1 -

u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[(1 - u)^2 - (1 - 2

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /

Rubi steps \begin{align*} \text {integral}& = -\left ((i c) \int \frac {(a+b \arctan (c x))^2}{x^2 (d+i c d x)} \, dx\right )+\frac {\int \frac {(a+b \arctan (c x))^2}{x^3} \, dx}{d} \\ & = -\frac {(a+b \arctan (c x))^2}{2 d x^2}-c^2 \int \frac {(a+b \arctan (c x))^2}{x (d+i c d x)} \, dx-\frac {(i c) \int \frac {(a+b \arctan (c x))^2}{x^2} \, dx}{d}+\frac {(b c) \int \frac {a+b \arctan (c x)}{x^2 \left (1+c^2 x^2\right )} \, dx}{d} \\ & = -\frac {(a+b \arctan (c x))^2}{2 d x^2}+\frac {i c (a+b \arctan (c x))^2}{d x}-\frac {c^2 (a+b \arctan (c x))^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {(b c) \int \frac {a+b \arctan (c x)}{x^2} \, dx}{d}-\frac {\left (2 i b c^2\right ) \int \frac {a+b \arctan (c x)}{x \left (1+c^2 x^2\right )} \, dx}{d}-\frac {\left (b c^3\right ) \int \frac {a+b \arctan (c x)}{1+c^2 x^2} \, dx}{d}+\frac {\left (2 b c^3\right ) \int \frac {(a+b \arctan (c x)) \log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d} \\ & = -\frac {b c (a+b \arctan (c x))}{d x}-\frac {3 c^2 (a+b \arctan (c x))^2}{2 d}-\frac {(a+b \arctan (c x))^2}{2 d x^2}+\frac {i c (a+b \arctan (c x))^2}{d x}-\frac {c^2 (a+b \arctan (c x))^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {i b c^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{d}+\frac {\left (2 b c^2\right ) \int \frac {a+b \arctan (c x)}{x (i+c x)} \, dx}{d}+\frac {\left (b^2 c^2\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx}{d}+\frac {\left (i b^2 c^3\right ) \int \frac {\operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d} \\ & = -\frac {b c (a+b \arctan (c x))}{d x}-\frac {3 c^2 (a+b \arctan (c x))^2}{2 d}-\frac {(a+b \arctan (c x))^2}{2 d x^2}+\frac {i c (a+b \arctan (c x))^2}{d x}-\frac {2 i b c^2 (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )}{d}-\frac {c^2 (a+b \arctan (c x))^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {i b c^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{d}-\frac {b^2 c^2 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d}+\frac {\left (b^2 c^2\right ) \text {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )}{2 d}+\frac {\left (2 i b^2 c^3\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{d} \\ & = -\frac {b c (a+b \arctan (c x))}{d x}-\frac {3 c^2 (a+b \arctan (c x))^2}{2 d}-\frac {(a+b \arctan (c x))^2}{2 d x^2}+\frac {i c (a+b \arctan (c x))^2}{d x}-\frac {2 i b c^2 (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )}{d}-\frac {c^2 (a+b \arctan (c x))^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {b^2 c^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )}{d}-\frac {i b c^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{d}-\frac {b^2 c^2 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d}+\frac {\left (b^2 c^2\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d}-\frac {\left (b^2 c^4\right ) \text {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )}{2 d} \\ & = -\frac {b c (a+b \arctan (c x))}{d x}-\frac {3 c^2 (a+b \arctan (c x))^2}{2 d}-\frac {(a+b \arctan (c x))^2}{2 d x^2}+\frac {i c (a+b \arctan (c x))^2}{d x}+\frac {b^2 c^2 \log (x)}{d}-\frac {b^2 c^2 \log \left (1+c^2 x^2\right )}{2 d}-\frac {2 i b c^2 (a+b \arctan (c x)) \log \left (2-\frac {2}{1-i c x}\right )}{d}-\frac {c^2 (a+b \arctan (c x))^2 \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {b^2 c^2 \operatorname {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )}{d}-\frac {i b c^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{d}-\frac {b^2 c^2 \operatorname {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d} \\ \end{align*}

Mathematica [A] (veriﬁed)

Time = 1.23 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.37 \[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx=\frac {-\frac {a^2}{x^2}+\frac {2 i a^2 c}{x}+2 i a^2 c^2 \arctan (c x)-2 a^2 c^2 \log (x)+a^2 c^2 \log \left (1+c^2 x^2\right )+\frac {2 i a b \left (2 c^2 x^2 \arctan (c x)^2+\arctan (c x) \left (i+2 c x+i c^2 x^2+2 i c^2 x^2 \log \left (1-e^{2 i \arctan (c x)}\right )\right )+c x \left (i-2 c x \log (c x)+c x \log \left (1+c^2 x^2\right )\right )+c^2 x^2 \operatorname {PolyLog}\left (2,e^{2 i \arctan (c x)}\right )\right )}{x^2}+2 b^2 c^2 \left (\frac {i \pi ^3}{24}-\frac {\arctan (c x)}{c x}-\frac {3}{2} \arctan (c x)^2-\frac {\arctan (c x)^2}{2 c^2 x^2}+\frac {i \arctan (c x)^2}{c x}-\arctan (c x)^2 \log \left (1-e^{-2 i \arctan (c x)}\right )-2 i \arctan (c x) \log \left (1-e^{2 i \arctan (c x)}\right )+\log (c x)-\frac {1}{2} \log \left (1+c^2 x^2\right )-i \arctan (c x) \operatorname {PolyLog}\left (2,e^{-2 i \arctan (c x)}\right )-\operatorname {PolyLog}\left (2,e^{2 i \arctan (c x)}\right )-\frac {1}{2} \operatorname {PolyLog}\left (3,e^{-2 i \arctan (c x)}\right )\right )}{2 d} \]

Integrate[(a + b*ArcTan[c*x])^2/(x^3*(d + I*c*d*x)),x]

(-(a^2/x^2) + ((2*I)*a^2*c)/x + (2*I)*a^2*c^2*ArcTan[c*x] - 2*a^2*c^2*Log[x] + a^2*c^2*Log[1 + c^2*x^2] + ((2*

I)*a*b*(2*c^2*x^2*ArcTan[c*x]^2 + ArcTan[c*x]*(I + 2*c*x + I*c^2*x^2 + (2*I)*c^2*x^2*Log[1 - E^((2*I)*ArcTan[c

*x])]) + c*x*(I - 2*c*x*Log[c*x] + c*x*Log[1 + c^2*x^2]) + c^2*x^2*PolyLog[2, E^((2*I)*ArcTan[c*x])]))/x^2 + 2

*b^2*c^2*((I/24)*Pi^3 - ArcTan[c*x]/(c*x) - (3*ArcTan[c*x]^2)/2 - ArcTan[c*x]^2/(2*c^2*x^2) + (I*ArcTan[c*x]^2

)/(c*x) - ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] - (2*I)*ArcTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] + L

og[c*x] - Log[1 + c^2*x^2]/2 - I*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] - PolyLog[2, E^((2*I)*ArcTan[c

*x])] - PolyLog[3, E^((-2*I)*ArcTan[c*x])]/2))/(2*d)

Maple [C] (warning: unable to verify)

Time = 30.95 (sec) , antiderivative size = 1873, normalized size of antiderivative = 6.86


method	result	size

derivativedivides	\(\text {Expression too large to display}\)	\(1873\)
default	\(\text {Expression too large to display}\)	\(1873\)
parts	\(\text {Expression too large to display}\)	\(1877\)

int((a+b*arctan(c*x))^2/x^3/(d+I*c*d*x),x,method=_RETURNVERBOSE)

c^2*(-1/2*a^2/d/c^2/x^2+I*a^2/d/c/x-a^2/d*ln(c*x)+1/2*a^2/d*ln(c^2*x^2+1)+I*a^2/d*arctan(c*x)+b^2/d*(-1/2/c^2/

x^2*arctan(c*x)^2-2*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))+ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*polylog(3,-(1+I

*c*x)/(c^2*x^2+1)^(1/2))+ln((1+I*c*x)/(c^2*x^2+1)^(1/2)-1)-3/2*arctan(c*x)^2+arctan(c*x)^2*ln((1+I*c*x)^2/(c^2

*x^2+1)-1)-arctan(c*x)^2*ln(c*x)-arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))-arctan(c*x)^2*ln(1+(1+I*c*x)/

(c^2*x^2+1)^(1/2))-2*dilog(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*dilog((1+I*c*x)/(c^2*x^2+1)^(1/2))-1/2*I*Pi*csgn(I

/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c

*x)^2/(c^2*x^2+1)))*arctan(c*x)^2+2*I*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*I*arctan(c*x)*polyl

og(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2))+1/2*I*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*cs

gn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^2-1/2*I*Pi*csgn(I*((1+I*c*x)^2/(c^2*

x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))*arctan(c*

x)^2+1/2*I*Pi*csgn(I/(1+(1+I*c*x)^2/(c^2*x^2+1)))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1

)))^2*arctan(c*x)^2+1/2*I*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*

x)^2/(c^2*x^2+1)))^2*arctan(c*x)^2-1/2*I*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^3*ar

ctan(c*x)^2+1/2*I*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^2-1/2*I*Pi*cs

gn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^3*arctan(c*x)^2-1/2*arctan(c*x)*(I*c*x-(c^2*x^2+

1)^(1/2)+1)/c/x-1/2*arctan(c*x)*(I*c*x+(c^2*x^2+1)^(1/2)+1)/c/x+I*arctan(c*x)^2/c/x-3/2*I*Pi*arctan(c*x)^2-arc

tan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))+arctan(c*x)^2*ln(c*x-I)+2/3*I*arctan(c*x)^3+1/2*I*Pi*csgn((1+I*c*x)

^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^2-1/2*I*Pi*csgn(I/(1+(

1+I*c*x)^2/(c^2*x^2+1)))*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan(c*x)^2-2*I*arctan(

c*x)*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+I*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^2*arctan

(c*x)^2+1/2*I*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1)))^3*arctan(c*x)^2+1/2*I*Pi*csgn(I/(1+

(1+I*c*x)^2/(c^2*x^2+1)))*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/(1+(1+I*c*x)^2/(c^2*x^2+1

)))*arctan(c*x)^2)+2/d*a*b*(-1/2/c^2/x^2*arctan(c*x)+I*arctan(c*x)/c/x-arctan(c*x)*ln(c*x)+arctan(c*x)*ln(c*x-

I)-I*ln(c*x)-1/2/c/x+1/2*I*ln(c^2*x^2+1)-1/2*arctan(c*x)-1/2*I*ln(c*x)*ln(1+I*c*x)+1/2*I*ln(c*x)*ln(1-I*c*x)-1

/2*I*dilog(1+I*c*x)+1/2*I*dilog(1-I*c*x)-1/2*I*(dilog(-1/2*I*(c*x+I))+ln(c*x-I)*ln(-1/2*I*(c*x+I)))+1/4*I*ln(c

Fricas [F]

\[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )} x^{3}} \,d x } \]

integrate((a+b*arctan(c*x))^2/x^3/(d+I*c*d*x),x, algorithm="fricas")

1/8*(2*b^2*c^2*x^2*log(2*c*x/(c*x - I))*log(-(c*x + I)/(c*x - I))^2 + 4*b^2*c^2*x^2*dilog(-2*c*x/(c*x - I) + 1

)*log(-(c*x + I)/(c*x - I)) - 4*b^2*c^2*x^2*polylog(3, -(c*x + I)/(c*x - I)) + 8*d*x^2*integral(1/2*(-2*I*a^2*

c*x + 2*a^2 + (2*b^2*c^2*x^2 + (2*a*b + I*b^2)*c*x + 2*I*a*b)*log(-(c*x + I)/(c*x - I)))/(c^2*d*x^5 + d*x^3),

x) + (-2*I*b^2*c*x + b^2)*log(-(c*x + I)/(c*x - I))^2)/(d*x^2)

Sympy [F]

\[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx=- \frac {i \left (\int \frac {a^{2}}{c x^{4} - i x^{3}}\, dx + \int \frac {b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{c x^{4} - i x^{3}}\, dx + \int \frac {2 a b \operatorname {atan}{\left (c x \right )}}{c x^{4} - i x^{3}}\, dx\right )}{d} \]

integrate((a+b*atan(c*x))**2/x**3/(d+I*c*d*x),x)

-I*(Integral(a**2/(c*x**4 - I*x**3), x) + Integral(b**2*atan(c*x)**2/(c*x**4 - I*x**3), x) + Integral(2*a*b*at

Maxima [F]

\[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )} x^{3}} \,d x } \]

integrate((a+b*arctan(c*x))^2/x^3/(d+I*c*d*x),x, algorithm="maxima")

1/2*(2*c^2*log(I*c*x + 1)/d - 2*c^2*log(x)/d + (2*I*c*x - 1)/(d*x^2))*a^2 - 1/96*(-24*I*b^2*c^2*x^2*arctan(c*x

)^3 + 3*b^2*c^2*x^2*log(c^2*x^2 + 1)^3 - 2*(384*b^2*c^4*integrate(1/16*x^4*arctan(c*x)^2/(c^2*d*x^5 + d*x^3),

x) + b^2*c^2*log(c^2*x^2 + 1)^3/d + 12*b^2*c^2*arctan(c*x)^2/d + 96*b^2*c^2*integrate(1/16*x^2*log(c^2*x^2 + 1

)/(c^2*d*x^5 + d*x^3), x) - 192*b^2*c*integrate(1/16*x*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*d*x^5 + d*x^3), x) +

192*b^2*c*integrate(1/16*x*arctan(c*x)/(c^2*d*x^5 + d*x^3), x) + 576*b^2*integrate(1/16*arctan(c*x)^2/(c^2*d*x

^5 + d*x^3), x) + 48*b^2*integrate(1/16*log(c^2*x^2 + 1)^2/(c^2*d*x^5 + d*x^3), x) + 1536*a*b*integrate(1/16*a

rctan(c*x)/(c^2*d*x^5 + d*x^3), x))*d*x^2 + 16*I*(b^2*c^2*arctan(c*x)^3/d + 12*b^2*c^3*integrate(1/16*x^3*log(

c^2*x^2 + 1)^2/(c^2*d*x^5 + d*x^3), x) - 24*b^2*c^3*integrate(1/16*x^3*log(c^2*x^2 + 1)/(c^2*d*x^5 + d*x^3), x

) + 24*b^2*c^2*integrate(1/16*x^2*arctan(c*x)/(c^2*d*x^5 + d*x^3), x) + 72*b^2*c*integrate(1/16*x*arctan(c*x)^

2/(c^2*d*x^5 + d*x^3), x) + 6*b^2*c*integrate(1/16*x*log(c^2*x^2 + 1)^2/(c^2*d*x^5 + d*x^3), x) + 192*a*b*c*in

tegrate(1/16*x*arctan(c*x)/(c^2*d*x^5 + d*x^3), x) - 12*b^2*c*integrate(1/16*x*log(c^2*x^2 + 1)/(c^2*d*x^5 + d

*x^3), x) + 24*b^2*integrate(1/16*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*d*x^5 + d*x^3), x))*d*x^2 + 12*(-2*I*b^2*c

*x + b^2)*arctan(c*x)^2 - 3*(2*I*b^2*c^2*x^2*arctan(c*x) - 2*I*b^2*c*x + b^2)*log(c^2*x^2 + 1)^2 + 12*(b^2*c^2

*x^2*arctan(c*x)^2 + (2*b^2*c*x + I*b^2)*arctan(c*x))*log(c^2*x^2 + 1))/(d*x^2)

Giac [F]

\[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )} x^{3}} \,d x } \]

integrate((a+b*arctan(c*x))^2/x^3/(d+I*c*d*x),x, algorithm="giac")

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx=\int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{x^3\,\left (d+c\,d\,x\,1{}\mathrm {i}\right )} \,d x \]

int((a + b*atan(c*x))^2/(x^3*(d + c*d*x*1i)),x)

int((a + b*atan(c*x))^2/(x^3*(d + c*d*x*1i)), x)

\(\int \frac {(a+b \arctan (c x))^2}{x^3 (d+i c d x)} \, dx\) [101]

Optimal result

Rubi [A] (veriﬁed)

Mathematica [A] (veriﬁed)

Maple [C] (warning: unable to verify)

Fricas [F]

Sympy [F]

Maxima [F]

Giac [F]

Mupad [F(-1)]